daurialebraslmc

## Answered question

2022-04-06

Finding:
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}$

### Answer & Explanation

Assorrymarf0cgr

Beginner2022-04-07Added 10 answers

Notice that:
${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}dx=2{\int }_{0}^{+\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}dx$
$=2{\int }_{0}^{1}\frac{{x}^{2}}{1+{x}^{4}}dx+2{\int }_{1}^{+\mathrm{\infty }}\frac{{x}^{2}}{1+{x}^{4}}dx$
$=2{\int }_{0}^{1}\frac{1+{x}^{2}}{1+{x}^{4}}dx$
$=2\left(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\dots \right)$
$=2\left(1+\frac{2}{{4}^{2}-1}-\frac{2}{{8}^{2}-1}+\frac{2}{{12}^{2}-1}-\dots \right)$
and, from the logarithmic derivative of the Weierstrass product for the sine and cosine function:
$\sum _{k\ge 0}\frac{1}{{\left(2k+1\right)}^{2}-{x}^{2}}=\frac{\pi }{4x}\mathrm{tan}\left(\frac{\pi x}{2}\right)$,
$\sum _{k\ge 1}\frac{1}{{k}^{2}-{x}^{2}}=\frac{1-\pi x\mathrm{cot}\left(\pi x\right)}{2{x}^{2}}$,
so by taking limits as $x\to \frac{1}{2}$ we get:
${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}dx=\frac{\pi }{\sqrt{2}}$

blessgansgxei

Beginner2022-04-08Added 10 answers

Notice
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}dx$
$=2{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}dx$
$=2{\int }_{0}^{\mathrm{\infty }}\frac{dx}{{x}^{2}+{x}^{-2}}$
$=2{\int }_{0}^{\mathrm{\infty }}\frac{dx}{{\left(x-{x}^{-1}\right)}^{2}+2}$
$=2\left({\int }_{0}^{1}+{\int }_{1}^{\mathrm{\infty }}\right)\frac{dx}{{\left(x-{x}^{-1}\right)}^{2}+2}$
Change variable from x to $\frac{1}{x}$ for that part of integral on $\left(0,1\right)$, this becomes
$2{\int }_{1}^{\mathrm{\infty }}\frac{1}{{\left(x-{x}^{-1}\right)}^{2}+2}\left(1+{x}^{-2}\right)dx=2{\int }_{1}^{\mathrm{\infty }}\frac{d\left(x-{x}^{-1}\right)}{{\left(x-{x}^{-1}\right)}^{2}+2}$
Change variable once more to $y=x-{x}^{-1}$ and then $y=\sqrt{2}z$, we get
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{x}^{2}}{{x}^{4}+1}dx=2{\int }_{0}^{\mathrm{\infty }}\frac{dy}{{y}^{2}+2}$
$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dy}{{y}^{2}+2}$
$=\frac{1}{\sqrt{2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dz}{1+{z}^{2}}$
$=\frac{\pi }{\sqrt{2}}$

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